# Predicting mean smooth in GAM with smooth-by-random-factor interaction

I have a binomial GAM with a smooth-by-random-factor interaction. From this I am able to predict and visualize the smooth term for any level of my random effects:

#Simulate data
set.seed(0)
means = rnorm(5, mean=0, sd=2)
df = data_frame(group = as.factor(rep(1:5, each=100)),
x = rep(seq(-3,3, length.out =100), 5),
y=as.numeric(dnorm(x, mean=means[group]) > 0.4*runif(10)))

#Fit model
library(mgcv)
gam_model = gam(y ~ te(x, group, bs=c("ts", "re")), data=df, family = binomial)

#Visualize
df2 = predict(gam_model, type="response", se.fit=TRUE)
df2 = cbind(df, response = df2$fit, lwr = df2$fit-2*df2$se.fit, upr = df2$fit+2*df2$se.fit) library(ggplot2) ggplot() + geom_ribbon(data = df2, mapping=aes(x=x, ymin=lwr, ymax=upr, fill=group), alpha=0.25) + geom_line(data = df2, mapping=aes(x=x, y=response, col=group)) + geom_point(data = df, mapping=aes(x=x, y=y, col=group)) + facet_wrap(~group) How do I predict the mean smooth and confidence intervals around it? ## 2 Answers The solution suggested by Simon Wood to the simpler problem of predicting the population level effect from a model with random intercepts represented as a smooth is to use a by variable in the random effect smooth. See this Answer for some detail. You can't do this dummy trick directly with your model as you have the smooth and random effects all bound up in the 2d spline term. As I understand it, you should be able to decompose your tensor product spline into "main effects" and the "spline interaction". I quote these as the decomposition will be to split out the fixed effects and random effects parts of the model. Nb: I think I have this right but it would be helpful to have people knowledgeable with mgcv give this a once over. ## load packages library("mgcv") library("ggplot2") set.seed(0) means <- rnorm(5, mean=0, sd=2) ## generate data df <- data.frame(group = as.factor(rep(1:5, each=100)), x = rep(seq(-3,3, length.out =100), 5), y = as.numeric(dnorm(x, mean=means[group]) > 0.4*runif(10)), dummy = 1) # dummy variable trick  This is what I came up with: gam_model3 <- gam(y ~ s(x, bs = "ts") + s(group, bs = "re", by = dummy) + ti(x, group, bs = c("ts","re"), by = dummy), data = df, family = binomial, method = "REML")  Here I've broken out the fixed effects smooth of x, the random intercepts and the random - smooth interaction. Each of the random effect terms includes by = dummy. This allows us to zero out these terms by switching dummy to be a vector of 0s. This works because by terms here multiply the smooth by a numeric value; where dummy == 1 we get the effect of the random effect smooth but when dummy == 0 we are multiplying the effect of each random effect smoother by 0. To get the population level we need just the effect of s(x, bs = "ts") and zero out the other terms. newdf <- data.frame(group = as.factor(rep(1, 100)), x = seq(-3, 3, length = 100), dummy = rep(0, 100)) # zero out ranef terms ilink <- family(gam_model3)$linkinv      # inverse link function
df2 <- predict(gam_model3, newdf, se.fit = TRUE)
ilink <- family(gam_model3)$linkinv df2 <- with(df2, data.frame(newdf, response = ilink(fit), lwr = ilink(fit - 2*se.fit), upr = ilink(fit + 2*se.fit)))  (Note that all this was done on the scale of the linear predictor and only backtransformed at the end using ilink()) Here's what the population-level effect looks like theme_set(theme_bw()) p <- ggplot(df2, aes(x = x, y = response)) + geom_point(data = df, aes(x = x, y = y, colour = group)) + geom_ribbon(aes(ymin = lwr, ymax = upr), alpha = 0.1) + geom_line() p And here are the group level smooths with the population level one superimposed df3 <- predict(gam_model3, se.fit = TRUE) df3 <- with(df3, data.frame(df, response = ilink(fit), lwr = ilink(fit - 2*se.fit), upr = ilink(fit + 2*se.fit)))  and a plot p2 <- ggplot(df3, aes(x = x, y = response)) + geom_point(data = df, aes(x = x, y = y, colour = group)) + geom_ribbon(aes(ymin = lwr, ymax = upr, fill = group), alpha = 0.1) + geom_line(aes(colour = group)) + geom_ribbon(data = df2, aes(ymin = lwr, ymax = upr), alpha = 0.1) + geom_line(data = df2, aes(y = response)) p2  From a cursory inspection this looks qualitatively similar to the result from Ben's answer but it is smoother; you don't get the blips where the next group's data is not all zero. It depends. There are a bunch of ways to define the "average" response. The answer here is based on the unweighted average across groups; with this simple, artificial example it doesn't make any difference, but in other cases you might want to take a population-weighted average. n.b. there are several reasons the following is not quite right, although it's not an unreasonable start • for consistency, we ought to be taking the mean / combining the standard errors on the linear predictor (link) scale, not the response scale • the answer below essentially treats the groups as fixed effects. We know more about the distribution (at least the assumed distribution) of the conditional modes ... but it means there are a lot of possible definitions I will update when I get a chance, but it's still a slightly useful answer Repeating data generation for my convenience ... library("dplyr") ## for data_frame set.seed(0) means = rnorm(5, mean=0, sd=2) df = data_frame(group = as.factor(rep(1:5, each=100)), x = rep(seq(-3,3, length.out =100), 5), y=as.numeric(dnorm(x, mean=means[group]) > 0.4*runif(10))) #Fit model library(mgcv) gam_model = gam(y ~ te(x, group, bs=c("ts", "re")), data=df, family = binomial) gam_avg = gam(y ~ s(x), data=df, family = binomial)  Tweak prediction step a tiny bit to retain se in the results (it would be nice to write a broom::augment() method for this case ...) #Predict pfun <- function(x, type="response") { pp <- predict(x, type="response", se.fit=TRUE) df2 = predict(gam_model, type="response", se.fit=TRUE) df2 = with(df2,data.frame(df, response = fit, se= se.fit, lwr = fit-2*se.fit, upr = fit+2*se.fit))  Generate mean predictions by averaging at each value of x; construct confidence intervals by adding "in quadrature" (i.e. sqrt(sum(x^2))) (I don't know why c() is necessary, but it seems to be). sumquad <- function(x) { sqrt(sum(c(x)^2)) } dfsum <- df2 %>% group_by(x) %>% summarise(response=mean(c(response)), se=sumquad(se)) %>% mutate(lwr=response-2*se,upr=response+2*se)  Now visualize: library("ggplot2"); theme_set(theme_bw()) gg1 <- ggplot(mapping=aes(x)) + geom_ribbon(data = df2, mapping=aes(ymin=lwr, ymax=upr, fill=group), alpha=0.25) + geom_line(data = df2, mapping=aes(y=response, col=group)) + geom_point(data = df, mapping=aes(y=y, col=group)) ## add mean response + ribbon gg1 + geom_line(data=dfsum,aes(y=response))+ geom_ribbon(data=dfsum,aes(ymin=lwr,ymax=upr),alpha=0.2)  • To switch to the link scale for this answer, would it require more than the following? 1) changing to type="link" and 2) adding  mutate(response=plogis(link), re_upr=plogis(upr), re_lwr=plogis(lwr)) for both df2 and dfsum – jaimedash Jan 5 '16 at 19:08 • @jaimedash no (but plogis(response)), I don't think so (i.e. yes I think this is all you need). You're supposed to do all these things on the scale of the linear predictor and then back transform. More generally you can always grab the inverse of the link function via ilink <- family(mod)$linkinv and use ilink(foo). – Reinstate Monica - G. Simpson Jan 5 '16 at 19:32